Spring 2006
Process Dynamics, Operations, and Control
10.450
Lesson 4: Two Tanks in Series
4.0
context and direction
In Lesson 3 we performed a material balance on a mixing tank and derived
a first-order system model. We used that model to predict the open-loop
process behavior and its closed-loop behavior, under feedback control. In
this lesson, we complicate the process, and find that some additional
analysis tools will be useful.
DYNAMIC SYSTEM BEHAVIOR
4.1
math model of continuous blending tanks
We consider two tanks in series with single inlet and outlet streams.
F, CAi
F, CA1
F, CA2
volume V1
volume V2
Our component A mass balance is written over each tank.
d
V1C A1 = FC Ai − FC A1
dt
d
V2 C A 2 = FC A1 − FC A 2
dt
(4.1-1)
As in Lesson 3, we have recognized that each tank operates in overflow:
the volume is constant, so that changes in the inlet flow are quickly
duplicated in the outlet flow. Hence all streams are written in terms of a
single volumetric flow F. Again, we will regard the flow as constant in
time. Also, each tank is well mixed.
Putting (4.1-1) into standard form
dC A1
+ C A1 = C Ai
dt
dC A 2
τ2
+ C A 2 = C A1
dt
τ1
revised 2006 Mar 6
(4.1-2)
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Lesson 4: Two Tanks in Series
we identify two first-order dynamic systems coupled through the
composition of the intermediate stream, CA1. If we view the tanks as
separate systems, we see that CA1 is the response variable of the first tank
and the input to the second. If instead we view the pair of tanks as a single
system, CA1 becomes an intermediate variable. The speed of response
depends on two time constants, which (as before) are equal to the ratio of
volume for each tank and the common volumetric flow.
We write (4.1-2) at a steady reference condition to find
C A ,r = C Ai ,r = C A1,r = C A 2,r
(4.1-3)
We subtract the reference condition from (4.1-2) and thus express the
variables in deviation form.
dC 'A1
τ1
+ C 'A1 = C 'Ai
dt
dC 'A 2
τ2
+ C 'A 2 = C 'A1
dt
(4.1-4)
4.2
solving the coupled equations - a second-order system
As usual, we will take the initial condition to be zero (response variables
at their reference conditions). We may solve (4.1-4) in two ways:
Because the first equation contains only C′A1, we may integrate it directly
to find C′A1 as a function of the input C′Ai. This solution becomes the
forcing function in the second equation, which may be integrated directly
to find C′A2. That is
t
C'A1 =
1 − t τ1 t τ1 '
e ∫ e C Ai dt
τ1
0
(4.2-1)
C 'A 2 =
t
t
1 − t τ2 t τ 2 ⎡ 1 − t τ1 t τ1 ' ⎤
e ∫ e ⎢ e ∫ e C Ai dt ⎥ dt
τ2
0
0
⎣ τ1
⎦
(4.2-2)
On defining a specific disturbance C′Ai we can integrate (4.2-2) to a
solution.
Alternatively, we may eliminate the intermediate variable C′A1 between
the equations (4.1-4) and obtain a second-order equation for C′A2 as a
function of C′Ai. The steps are
(1) differentiate the second equation
(2) solve the first equation for the derivative of C′A1
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Lesson 4: Two Tanks in Series
(3) solve the original second equation for C′A1
(4) substitute in the equation of the first step.
The result is
τ1τ 2
d 2 C 'A 2
dC 'A 2
(
)
+
τ
+
τ
+ C 'A 2 = C 'Ai
1
2
dt 2
dt
(4.2-3)
Two mass storage elements led to two first-order equations, which have
combined to produce a single second-order equation. A homogeneous
solution to (4.2-3) can be found directly, but the particular solution
depends on the nature of the disturbance:
C'A 2 = A1e
−t
τ1
+ A 2e
−t
τ2
( )
+ C'A 2,part C 'Ai
(4.2-4)
where the constants A1 and A2 are found by invoking initial conditions
after the particular solution is determined.
4.3
response of system to step disturbance
Suppose a step change ΔC occurs in the inlet concentration at time td.
Either (4.2-2) or (4.2-4) yields
− (t − t d )
− (t − t d )
⎡
τ1
τ2
τ1
τ2 ⎤
C'A2 = U ( t − t d ) ΔC ⎢1 −
e
e
+
⎥
τ1 − τ2
⎣ τ1 − τ2
⎦
(4.3-1)
Each tank contributes a first-order response based on its own time
constant. However, these responses are weighted by factors that depend
on both time constants.
The result in Figure 4.3-1 looks somewhat different from the first-order
responses we have seen. We have plotted the step response of a secondorder system with τ1 = 1 and τ1 = 1.5 in arbitrary units. At sufficiently
long time, the initial condition has no influence and the outlet
concentration will become equal to the new inlet concentration; in this
respect it looks like the first-order system response. However, the initial
behavior differs: the outlet concentration rises gradually instead of
abruptly. This S-shaped curve, often called “sigmoid”, is a feature of
systems of order greater than one. Physically, we can understand this by
realizing that the change in inlet concentration must spread through two
tanks, and it reaches the second tank only after being diluted in the first.
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Lesson 4: Two Tanks in Series
C* Ai/ C
1
0.5
0
1
0
1
2
3
4
5
6
0
1
2
3
4
5
6
C*A2/ C
0.8
0.6
0.4
0.2
0
t
Figure 4.3-1: Response to step change in inlet composition
4.4
introducing the Laplace transform
We bother with the Laplace transform for two reasons:
• after the initial learning pains, it actually makes the math easier, so we
will use it in derivations
• some of the terminology in linear systems and process control is based
on formulating the equations with Laplace transforms.
Definition: the Laplace transform turns a function of time y(t) into a
function of the complex variable s. Variable s has dimensions of
reciprocal time. All the information contained in the time-domain
function is preserved in the Laplace domain.
∞
y(s) = L{y( t )} = ∫ y( t )e −st dt
(4.4-1)
0
(In these notes, we use the notation y(s) merely to indicate that y(t) has
been transformed; we do not mean that y(s) has the same functional
dependence on s that it does on t.)
Functional transforms: textbooks (for example, Marlin, Sec. 4.2) usually
include tables of transform pairs, so these derivations from definition
(4.4-1) are primarily to demonstrate how the tables came to be.
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Lesson 4: Two Tanks in Series
∞
L{CU ( t − t d )} = ∫ CU ( t − t d )e −st dt
0
∞
= C ∫ e −st dt
(4.4-2)
td
− C −st
e
s
Ce −st d
=
s
=
{
∞
td
∞
}
L Ce −at = ∫ Ce −at e −st dt
0
− C −(s + a ) t ∞
e
0
s+a
C
=
s+a
=
(4.4-3)
Operational transforms: this allows us to transform entire equations, not
just particular functions.
∞
df ( t ) −st
⎧ df ( t ) ⎫
L⎨
e dt
⎬=∫
⎩ dt ⎭ 0 dt
∞
d
⎡
⎤
= ∫ ⎢sf ( t )e −st + (fe −st )⎥ dt
dt
⎦
0 ⎣
∞
= s ∫ f ( t )e −st dt + f ( t )e −st
0
(4.4-4)
∞
0
= sL{f ( t )}− f (0)
⎞
⎫ ∞⎛ t
⎧t
L ⎨∫ f (ξ )dξ⎬ = ∫ ⎜⎜ ∫ f (ξ )dξ ⎟⎟e −st dt
⎭ 0⎝0
⎠
⎩0
t
∞
∞
⎞ ⎞
1 ⎛⎜ ⎛
1
−st
⎜
= ∫ f (t )e dt − ∫ d ⎜ ∫ f (ξ )dξ ⎟⎟e −st ⎟
⎟
s 0 ⎜⎝ ⎝ 0
s0
⎠ ⎠
⎞
1⎛
1
= L{f (t )} − ⎜⎜ ∫ f (ξ )dξ ⎟⎟e −st
s⎝0
s
⎠
t
∞
(4.4-5)
0
1
= L{f (t )}
s
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Lesson 4: Two Tanks in Series
Inverting transforms: use the tables to invert simple Laplace-domain
functions to their time-domain equivalents. To simplify the polynomial
functions often found in control engineering we may use partial fraction
expansion. The complicated ratio in (4.4-6) can be inverted if is expanded
into a series of simpler fractions.
N (s)
N (s)
C1
C2
C3
=
=
+
+ ... +
+ ...
n −1
n
n
m
D(s) (s − α1 ) (s − α 2 ) ... (s − α1 )
(s − α1 )
(s − α 2 ) m
(4.4-6)
In (4.4-6), α1 and α2 are repeated roots of the denominator. The inverse
transform of each term will involve an exponential function of the root αi.
f (s) =
C
(s − α) n
⇒ f (t) =
Ct n −1e αt
(n − 1)!
(4.4-7)
Variety in the values of the coefficients Ci comes from the numerator
function N(s).
how to write the expansion
Arrange the denominator so that the coefficient of each s is 1. If there are
no repeated roots, each root appears in one term.
C3
C1
C2
N(s)
=
+
+
(s − α1 )(s − α 2 )(s − α 3 ) (s − α1 ) (s − α 2 ) (s − α 3 )
(4.4-8)
If a root is repeated, it requires a term for each repetition.
C3
C1
C2
N(s)
=
+
+
2
2
(s − α1 ) (s − α 2 )
(s − α 2 )
(s − α1 )(s − α 2 )
(4.4-9)
Some roots may appear as complex conjugate pairs, so that, for example
α1 = a + jb
α 2 = a − jb
(4.4-10)
where j is the square root of -1.
how to solve for the coefficients - it’s only algebra
1) For each of the real, distinct roots, multiply the expansion by each RH
denominator and substitute the value of the root for s to isolate the
coefficient. This also works for the highest power of a repeated root.
2) With some coefficients determined, it may be easiest to substitute
arbitrary values for s to get equations in the unknown coefficients.
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Lesson 4: Two Tanks in Series
3) For repeated roots, either
(3a) multiply the expansion by s and take the limit as s → ∞.
However, this will not isolate coefficients associated with
repeated complex roots.
(3b) multiply the expansion by the RH denominator of highest power.
Differentiate this equation with respect to s, and substitute the
value of the root for s. Continue differentiating in this manner to
isolate successive coefficients.
4) For complex roots, solving for one coefficient is enough. The other
coefficient will be the complex conjugate.
4.5
solving linear ODEs with Laplace transforms
We return to (4.1-4), the two equations that describe concentration in the
tanks.
dC 'A1
τ1
+ C 'A1 = C 'Ai
dt
dC 'A 2
τ2
+ C 'A 2 = C 'A1
dt
(4.1-4)
We perform the Laplace transform on the entire first equation. It
distributes across addition, and constant τ1 may be factored out.
⎧ dC '
⎫
L ⎨τ1 A1 + C 'A1 ⎬ = L C 'Ai
dt
⎩
⎭
⎧ dC ' ⎫
τ1L ⎨ A1 ⎬ + L C 'A1 = L C 'Ai
⎩ dt ⎭
{ }
{ } { }
(4.5-1)
We next perform an operational transform on the derivative. Because the
functional forms of the variables C′A1 and C′Ai are not yet known, we
simply indicate a variable in the Laplace domain.
{
}
τ1 sC'A1 (s) − C'A1 (0) + C'A1 (s) = C'Ai (s)
τ1sC'A1 (s) + C'A1 (s) = C'Ai (s)
(4.5-2)
We can easily solve (4.5-2) for C′A1. If we similarly treat the second
equation in (4.1-4), we arrive at the equivalent formulation in the Laplace
domain.
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Lesson 4: Two Tanks in Series
C 'A1 (s) =
1
C 'Ai (s)
τ1s + 1
1
C (s) =
C 'A1 (s)
τ 2s + 1
(4.5-3)
'
A2
Equations (4.5-3) are not solutions - we have not solved anything! We
merely have a new formulation of problem (4.1-4), a formulation that is
more abstract (what on earth is Laplace domain?) and yet simpler, by
virtue of being algebraic. It is important to remember that all the
information held in the differential equations (4.1-4) is preserved in the
Laplace domain formulation (4.5-3).
We proceed toward solution by eliminating the intermediate variable in
(4.5-3). We find
C 'A 2 (s) =
1
1
C 'Ai (s)
τ 2s + 1 τ1s + 1
(4.5-4)
With (4.5-4) we have gone as far as we can without knowing more about
the disturbance. That is, we cannot invert the right-hand side of (4.5-4)
until we can actually substitute a functional transform for the variable
C′Ai(s). In this sense, (4.5-4) resembles (4.2-2) and (4.2-4): a solution
needing more specification.
As in Section 4.3, suppose a step change ΔC occurs in the inlet
concentration at time td.
C 'Ai ( t ) = U (t − t d )ΔC
(4.5-5)
We must take the Laplace transform,
C 'Ai (s) =
ΔC − t d s
e
s
(4.5-6)
which we may substitute into (4.5-4).
C 'A 2 (s) =
1
1 ΔC − t ds
e
τ 2s + 1 τ1s + 1 s
(4.5-7)
This IS the solution, the step response of the two tanks in series. Of
course, it really must be inverted to the time domain. We treat the
polynomial denominator from either the tables or partial fraction
expansion:
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Lesson 4: Two Tanks in Series
−t
−t
⎧ 1
τ1
τ2
1 1⎫
τ1
τ2
=
−
+
L−1 ⎨
1
e
e
⎬
τ
+
τ
+
τ
−
τ
τ
−
τ
s
1
s
1
s
1
2
1
1
2
⎩ 2
⎭
(4.5-8)
Next we apply the time delay
−( t − t d )
−( t − t d )
⎧ 1
⎡
τ1
τ2
1 1 − t ds ⎫
τ1
τ2 ⎤
+
L−1 ⎨
e ⎬ = U(t − t d )⎢1 −
e
e
⎥
τ1 − τ 2
⎩ τ 2s + 1 τ1s + 1 s
⎣ τ1 − τ 2
⎦
⎭
(4.5-9)
Remembering the constant factor, we complete the inverse transform of
(4.5-7).
−( t − t d )
−( t − t d )
⎡
τ1
τ2
τ1
τ2 ⎤
+
C'A 2 = U( t − t d )ΔC⎢1 −
e
e
⎥
τ1 − τ 2
⎣ τ1 − τ 2
⎦
(4.5-10)
which, of course, is identical to (4.3-1), derived in the time domain.
4.6
describing systems with transfer functions
In Section 4.1 we derived a system model to describe transient behavior in a tanks-in-series
process. Then in Section 4.5 we used Laplace transforms to solve it. Let us now do the same
procedure in the abstract. Begin with the first-order lag, written in deviation variables:
τ
dy '
+ y ' = Kx ' ( t )
dt
y ' (0) = 0
(4.6-1)
After taking Laplace transforms, we relate input and output by an algebraic equation:
y ' (s) =
K '
x (s)
τs + 1
(4.6-2)
The ratio in (4.6-2) multiplies x′(s) (the transform of disturbance x′(t)) and in the process
converts that signal into y′(s) (the transform of the response y′(t)). We call this ratio the transfer
function G(s).
y ' (s)
K
G (s) = ' =
x (s) τs + 1
(4.6-3)
G(s) contains all the information about the ODE (4.6-1). We should from now recognize it,
when we see it, as a first-order lag. Should we want to know how the first-order lag behaves in
response to some disturbance, we transform the disturbance, multiply it by the first-order lag
transfer function, and then take the inverse transform of the result.
Let us generalize (4.5-4), which described two first-order lags in series:
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Lesson 4: Two Tanks in Series
y ' (s) =
K1
K2
x ' (s)
τ1s + 1 τ 2s + 1
(4.6-4)
The transfer function for this second-order system is a product of two firstorder lags
y ' (s)
K1
K2
=
= G1 (s)G 2 (s) = G (s)
'
x (s) τ1s + 1 τ 2s + 1
(4.6-5)
We shall consider a transfer function to be a completely satisfactory
description of a dynamic system. We shall learn to notice its gain (longterm steady-state relationship between y′ and x′), time constants, and poles
(roots of the denominator).
Table 4.6-1: Characteristics of systems we have studied
type
equation
transfer
function
1st order lag
K
dy'
+ y ' ( t ) = Kx ' ( t )
τs + 1
dt
nd
'
2
'
2 order
K
d y
dy
+ y ' = Kx ' ( t )
overdamped* τ1τ2 2 + (τ1 + τ2 )
( τ1s + 1)(τ 2s + 1)
dt
dt
nd
*why “overdamped”? There are other 2 order forms to be encountered.
τ
poles
steady
state
gain
-τ-1
K
-τ1-1
-τ2-1
K
4.7
describing systems with block diagrams
The block diagram is a graphical display of the system in the Laplace domain.
x′(s)
y′(s)
G(s)
It comprises blocks and arrows, and thus resembles many other types of flow diagram. In our
use with control systems, however, the arrows represent signals, variables that change in time,
which are not necessarily actual flow streams. The block contains the transfer function, which
may be as simple as a units conversion between x and y, or a description of more complicated
dynamic behavior. Remember that the transfer function incorporates all the dynamic
information in the system equations. This diagram implies the Laplace domain relationship
y ' (s) = G (s) x ' (s)
(4.7-1)
The real value of block diagrams is to represent the flow of signals among multiple blocks.
The Block Diagram Rules (see Marlin, Sec.4.4):
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Lesson 4: Two Tanks in Series
•
exclusion: only one input and output to a block.
inputs
•
summing: two signals may be summed at an explicit summing
junction. The algebraic sign is indicated at the junction (if omitted,
is presumed to be positive).
x'‘ 1(s)
G1(s)
y‘' 1(s)
+ x' 3(s)
x' 2(s)
•
outputs
system
G2(s)
G3(s)
y' 3(s)
y' 2(s)
multiple assignment: a single signal may feed its value to multiple
blocks. This does NOT indicate that the signal is divided up
among the blocks.
x 1(s)
x'
G1(s)
y 1(s)
y'
G2(s)
y' 1(s)
G3(s)
y 2(s)
y'
y' 3(s)
Block diagrams may be turned into equations by simple algebra. It is usually most convenient to
start with an output and work backwards by substitution. In the summing diagram
y 3' (s) = G 3 (s)x 3' (s)
= G 3 (s)(y1' (s) − y '2 (s) )
= G 3 (s)(G1 (s)x1' (s) − G 2 (s)x '2 (s) )
(4.7-2)
= G 3 (s)G1 (s)x1' (s) − G 3 (s)G 2 (s)x '2 (s)
In the multiple assignment diagram
y '2 (s) = G 2 (s) y1' (s)
= G 2 (s)G1 (s) x1' (s)
y 3' (s) = G 3 (s) y1' (s)
(4.7-3)
= G 3 (s)G1 (s) x1' (s)
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Similarly, equations may be turned into block diagrams. System (4.7-2) has two inputs and thus
requires at least 2 blocks.
x' 1(s)
G1(s) G3(s)
+ y' 3(s)
x' 2(s)
G2(s) G3(s)
System (4.7-3) has two outputs for one input. Input x*1 is not split – its full value is sent to each
of two blocks.
G1(s) G2(s)
x 1(s)
x'
G1(s) G3(s)
y' 2(s)
y 3(s)
y'
This pair of block diagrams is equivalent to the pair from which they were derived.
As a further illustration, we apply the block diagram rules to the two-tank
system in (4.5-3):
C' Ai(s)
C' A1(s)
C' A2(s)
1
1
τ1s + 1
τ 2s + 1
OR
C' Ai(s)
C' A2(s)
1
1
τ1s + 1 τ 2 s + 1
4.8
frequency response from the transfer function
In Section 3.5 we derived the system response to a sine input by
integrating the differential equation. We learned that the frequency
response - that is, the long-term oscillation - could be characterized by its
amplitude ratio and phase angle; these quantities were expressed on a
Bode plot.
Alternatively, we may derive the frequency response directly from the
transfer function by substituting jω for s, where j is the square root of -1
and ω is the radian frequency of the sine input. For the second-order
transfer function (4.6-5),
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Lesson 4: Two Tanks in Series
G ( jω) =
K 1K 2
(1 + τ1ωj)(1 + τ2ωj)
(1 − τ1ωj)(1 − τ 2ωj)
= K 1K 2
(1 + τ ω )(1 + τ
2
= K 1K 2
2
2
1
2
ω2
)
(4.8-1)
1 − τ1τ 2 ω2 − ω(τ1 + τ 2 ) j
2
2
1 + τ1 ω2 1 + τ 2 ω2
(
)(
)
The function G(jω), although perhaps daunting to behold, is simply a
complex number. As such, it has real and imaginary parts, and a
magnitude and a phase angle. It turns out that the magnitude of G(jω) is
the amplitude ratio of the frequency response.
(1 − τ τ ω ) + (ω(τ
2 2
G ( jω) = K1K 2
1 2
1
2
2
(1 + τ ω )(1 + τ ω )
1 + ω (τ + τ ) + ω τ τ
(1 + τ ω )(1 + τ ω )
2
2
2
= K 1K 2
2
2
1
=
+ τ 2 ))
2
1
2 2
1
2
2
4
2
2
1
2
2
2
(4.8-2)
2
K 1K 2
2
2
1 + τ1 ω2 1 + τ 2 ω2
Furthermore, the phase angle of G(jω) is the frequency response phase
angle.
Im(G ( jω) )
Re(G ( jω) )
− ω(τ1 + τ 2 )
= tan −1
1 − τ1τ 2 ω2
∠G ( jω) = tan −1
(4.8-3)
In Figure 4.8-1, the Bode plot abscissa has been normalized by the square
root of the product of the time constants. We see that the amplitude ratio
of a second-order system declines more swiftly than that of a first-order
system: the slope of the high-frequency asymptote is -2. Unbalancing the
time constants further decreases the amplitude ratio. The phase angle can
reach -180º. It is symmetric about -90º.
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Lesson 4: Two Tanks in Series
amplitude ratio/gain
1.00
0.10
0.01
0.01
0
1
1.1
-30
phase angle (deg)
0.1
τ2
= 20
τ1
-60
10
100
10
100
5
-90
-120
-150
-180
0.01
0.1
1
ω(τ1τ2)0.5 (radians)
Figure 4.8-1: Bode plot for overdamped second order system
4.9
stability of the two-tank system, and linear systems in general
The step response (4.5-10) shows no terms that grow with time, so long as
the time constants are positive. Furthermore, the amplitude ratio in Figure
4.8-1 is bounded. Thus, a second-order overdamped system appears to be
stable to bounded inputs. In practical terms, a concentration disturbance at
the inlet should not provoke a runaway response at the outlet.
We have seen first- and second-order linear systems. Let us generalize to
arbitrary order:
an
d n y'
d n −1 y '
dy '
"
+
a
+
+
a
+ y' = x '
n −1
1
dt
dt n
dt n −1
(4.9-1)
The function x′ represents all manner of bounded disturbances, expressed
in deviation form. The system properties, however, reside on the left-hand
side of (4.9-1), and its stability behavior should be independent of the
particular nature of the disturbances x′. Hence, we may examine the
homogeneous equation
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Lesson 4: Two Tanks in Series
an
d n y'
d n −1 y '
dy '
+
+
"
+
+ y' = 0
a
a
n −1
1
n
n −1
dt
dt
dt
(4.9-2)
The solution to (4.9-2) is the sum of n terms, each containing a factor eαit,
where αi is a real root, or the real part of a complex root, of the
characteristic equation.
a n r n + a n −1r n −1 + " + a 1r + 1 = 0
(4.9-3)
Hence if all the αi are negative, the solution cannot grow with time and
will thus be stable. If we take the Laplace transform of (4.9-1),
1
y ' (s)
= G (s) =
'
n
n −1
a n s + a n −1s + " + a 1s + 1
x (s)
(4.9-4)
we see that the denominator of the transfer function is identical to the
characteristic equation. Hence, the stability of the system is determined
by the poles of the transfer function: poles with zero or positive real parts
indicate a system unstable to bounded disturbances. Table 4.6-1 shows
that first-order lag and overdamped second-order systems are stable if
their time constants are positive.
CONTROL SCHEME
4.10 step 1 - specify a control objective for the process
Our control objective is to maintain the outlet composition CA2 at a
constant value.
4.11 step 2 - assign variables in the dynamic system
The controlled variable is clearly CA2. The inlet composition CAi is a
disturbance variable. The composition CA1 is an intermediate variable.
We have no candidate manipulated variable; hence we decide to add a
concentrated make-up stream as we did with the single tank in Lesson 3.
We could add the stream to the first or the second tank - we seek advice:
One advisor says, “Add it to the first tank, where the disturbance enters.
That way, the manipulation can interact thoroughly with the disturbance;
it’s a matter of success through cooperation.” A second advisor says,
“Add it to the second tank. That way the manipulated variable can affect
the controlled variable more directly, through one tank instead of two.”
This advice brings to mind the difference we have seen between first- and
second-order responses. Yet the first advisor is better-dressed, friendlier,
and has a comforting manner - we decide to add the make-up stream to the
first tank.
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Material balances on the solute give
V1 dC A1
F
C
+ C A1 =
C Ai + Ac Fc
F + Fc dt
F + Fc
F + Fc
(4.11-1)
V2 dC A 2
+ C A 2 = C A1
F + Fc dt
We expect to use a relatively small make-up flow Fc of concentration CAc;
hence we make the approximation that F + Fc ~ F. Hence
C
V1 dC A1
+ C A1 = C Ai + Ac Fc
F
F dt
V2 dC A 2
+ C A 2 = C A1
F dt
(4.11-2)
As is our custom, we write (4.11-2) at a steady reference condition and
subtract this reference to leave the variables in deviation form. After
taking Laplace transforms, we eliminate intermediate variable C′A1(s) to
find
C 'A 2 (s) =
1
(τ1s + 1)(τ 2s + 1)
C 'Ai (s) +
C Ac F
Fc' (s)
(τ1s + 1)(τ 2s + 1)
(4.11-3)
The time constants are the usual volume-to-flow ratios. The poles of the
two transfer functions are negative, so adding a make-up stream has not
made the system unstable. Compare (4.11-3) to (4.5-4), which describes
the two tanks without makeup flow. Figure 4.11-1, the block diagram of
(4.11-3), emphasizes that controlled variable CA2 is influenced by both
disturbance CAi and manipulated variable Fc.
C'Ai (s)
Fc' (s )
1
(τ1s + 1)(τ 2s + 1)
C Ac F
τ
+
( 1s 1)(τ2s + 1)
C'A 2 (s)
Figure 4.11-1: Block diagram of two-tank mixing process
4.12 step 3 - proportional control
We will use proportional control. Although we recognize the
disadvantage of offset, as demonstrated in Lesson 3, we feel confident in
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our ability to manage it at an acceptable level by increasing the controller
gain. The proportional algorithm is
Fc − Fbias = K gain (C A 2,setpt − C A 2 )
(4.12-1)
Should the outlet composition fall below the set point, the error will
become positive. A positive controller gain Kgain will increase make-up
flow Fc above the bias value, which will act to increase the outlet
composition.
4.13 step 4 - choose set points and limits
As in Section 3.16, we identify any applicable safety limits, choose the
desired operating point, specify limits of tolerable variation about it, and
supply make-up flow in sufficient quantity to counteract the anticipated
disturbances.
EQUIPMENT
4.14 components of the feedback control loop
We should consider the equipment more realistically. Figure 4.14-1 is a
block diagram showing the four components found in most process control
applications: process, sensor, controller, and final control element. Each
block contains a transfer function that relates output to input. The signals
between blocks are Laplace transforms of deviation variables. We
recognize that the process will typically comprise multiple blocks for
disturbance and manipulated variable inputs. An example is the mixing
tank process that shown in Figure 4.11-1.
process
x’d(s)
x’m(s)
final
control
element
Gd(s)
y’c (s)
Gm(s)
sensor
Gv(s)
y’
Gs(s)
controller
co(s)
y’sp,e(s)
Gc(s)
Gsp(s)
ε’ (s)
-
y’s (s)
y’sp(s)
Figure 4.14-1: General block diagram of feedback control loop
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The output signal that we have designated as the controlled variable y′c is
measured by a sensor. In our mixing tank example, we need some
measurement that reliably indicates composition; depending on the nature
of the solution, it might be a chromatograph, conductivity meter,
spectrometer, densitometer, etc.
The controller subtracts the sensor signal y′s from the set point y′sp and
executes the controller algorithm Gc(s) on the error. The subtraction is
sometimes said to take place in the comparator.
The controller output y′co drives a final control element to produce
manipulated variable x′m. In chemical process industries this is most often
a valve in a pipe carrying some fluid stream, but it could also be a motor, a
heater control, etc.
4.15 transfer functions for loop components
Although we are not yet ready to address hardware, we can improve our
description of equipment behavior by posing transfer functions for each of
the closed loop components indicated in Figure 4.14-1. Notice that a
transfer function has two main parts: steady and dynamic.
• The steady part is the gain; gain indicates the magnitude of the effect
of the transfer function on the input signal, and it performs the units
conversion between input and output. In (4.11-3), the transfer function
that converts make-up flow Fc into outlet concentration CA2 has a gain
of CAc/F. The gain depends on these two process parameters, and the
units are chosen to be consistent with those of the input and output
signals. The other transfer function in (4.11-3) has a dimensionless
unity gain, independent of process parameters.
• The dynamic part is everything else - all the system time constants and
the functions of the Laplace variable s. The dynamic part
characterizes the way that an input signal is processed in time. In
(4.11-3), both transfer functions feature second-order dynamics.
sensor
Let us presume that the sensor is fast - really fast - so that negligible time
elapses between a change in the controlled variable yc and its
measurement ys. Then the transfer function is
y s' = K s y 'c
(4.15-1)
Being really fast means that the transfer function has NO dynamic part.
Such a transfer function indicates a “pure gain” process, one in which
changes in the input are “instantaneously” seen in the output. The
dimensions of the gain Ks are
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K s ( =)
controller _ in
controlled var iable
(4.15-2)
where the sensor is presumed to deliver the measurement to the controller
in suitable “controller_in” units.
controller
We see that a proportional controller is also a pure gain process between
error signal and controller response.
y 'co = K c ε '
(4.15-3)
where the error is conventionally defined with the set point as positive:
'
ε ' = y sp
− y s'
(4.15-4)
For now we will leave the controller signal dimensions unspecified.
However, we can be sure that the gain has dimensions of
K c ( =)
controller _ out
controller _ in
(4.15-5)
set point
The error signal has dimensions suitable for the controller, which implies
that ysp and ys have the same units. However, the operator might prefer to
have the set point expressed in the units of the controlled variable (socalled engineering units), which implies that ysp,e and yc have the same
units. The set point transfer function performs this unit conversion; it is a
pure gain process with gain identical to that of the sensor. That is
G sp = K s
(4.15-6)
final control element
The valve is a mechanical device that takes some time to move. We might
imagine that a valve can change more quickly than a large chemical
process vessel, and thus that for many control applications the valve
dynamics can be neglected. In our case, however, we will assume that the
valve operates with first-order dynamics, such that
x 'm =
Kv '
y co
τ vs + 1
(4.15-7)
The valve gain has dimensions of
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K v ( =)
flow
controller _ out
(4.15-8)
CLOSED LOOP BEHAVIOR
4.16 assembling the components into a closed loop
The closed loop block diagram in Figure 4.14-1 shows how the loop
components are arranged. The transfer functions for the various
components are defined in Section 4.15. We will now use the block
diagram rules to derive the equations for the closed loop, in three steps:
• in general, good for any application of Figure 4.14-1
• applying the choices we made in Section 4.15
• adapting the general nomenclature of Section 4.15 to the two-tank
mixing process
We begin with the controlled variable, which is the output of the closed
loop system, and work backward through the diagram until all paths are
traced and the inputs appear.
y 'c (s) = G d x 'd (s) + G m x 'm (s)
= G d x 'd (s) + G m G v G c ε ' (s)
'
'
= G d x 'd (s) + G m G v G c (G sp y sp
, e (s) − G s y c (s) )
(4.16-1)
At this point, we collect the controlled variable on the left-hand side.
'
y 'c (s)(1 + G m G v G c G s ) = G d x 'd (s) + G m G v G c G sp y sp
,e (s)
y 'c (s) =
G m G v G c G sp
Gd
'
y sp
(s)
x 'd (s) +
(1 + G m G v G c G s )
(1 + G m G v G c G s ) ,e
(4.16-2)
Equation (4.16-2) shows how a controlled variable responds to a
disturbance and set point inputs. It is derived from Figure 4.14-1 and
applies to any system that can be represented by the figure.
We now specialize (4.16-2) with transfer functions we defined in Section
4.15. These use the general nomenclature of Figure 4.14-1, but depend on
assumptions we made about fast sensors and first-order valves.
Kv
K cKs
Gd
τ vs + 1
'
'
'
y c (s) =
x d (s) +
y sp
,e (s)
⎞
⎞
⎛
⎛
Kv
Kv
⎜⎜1 + G m
⎜⎜1 + G m
K c K s ⎟⎟
K c K s ⎟⎟
τ vs + 1
τ vs + 1
⎠
⎠
⎝
⎝
Gm
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Equation (4.16-3) begins to show how the general transfer functions
become specific functions of the Laplace variable s. Now we further
specialize (4.16-3) to the two-tank problem by substituting the process
transfer functions and specific nomenclature from (4.11-3) or,
equivalently, Figure 4.11-1.
C Ac F
Kv
KcKs
(
(
τ1s + 1)(τ 2s + 1) τ vs + 1
τ1s + 1)(τ 2s + 1)
'
'
C 'A 2,sp (s)
C Ai (s) +
C A 2 (s) =
⎞
⎞
⎛
⎛
C Ac F
Kv
C Ac F
Kv
⎜⎜1 +
⎜⎜1 +
K c K s ⎟⎟
K c K s ⎟⎟
⎠
⎠
⎝ (τ1s + 1)(τ 2s + 1) τ vs + 1
⎝ (τ1s + 1)(τ 2s + 1) τ v s + 1
1
(4.16-4)
Beneath its apparent complexity, (4.16-4) simply tells how the outlet
concentration reacts to disturbances and to a set point input. This closedloop response is compared in Figure 4.16-1 to the open-loop response of
the process.
transfer function for disturbance
1
C 'A 2 (s) =
(τ1s + 1)(τ 2s + 1)
other inputs
C 'Ai (s) +
C Ac F
Fc' (s)
(τ1s + 1)(τ2s + 1)
C Ac F
Kv
K cKs
(
(
τ1s + 1)(τ 2s + 1) τ v s + 1
τ1s + 1)(τ 2s + 1)
'
'
C A 2 (s) =
C Ai (s) +
C 'A 2,sp (s)
⎞
⎞
⎛
⎛
C Ac F
Kv
C Ac F
Kv
⎜⎜1 +
⎜⎜1 +
K c K s ⎟⎟
K c K s ⎟⎟
⎠
⎠
⎝ (τ1s + 1)(τ2s + 1) τ vs + 1
⎝ (τ1s + 1)(τ 2s + 1) τ v s + 1
1
Figure 4.16-1: Comparing open- and closed-loop responses
It is clear that the disturbance response has a different character, because
the transfer function has changed. IF we made a good decision on control
algorithm, and IF we tune the controller properly, the closed-loop response
should be better.
4.17 some perspective on how we derived the closed-loop response
Remember what we did: we proposed a block diagram of feedback control
and derived the associated transfer functions between inputs and output.
Then we substituted the component transfer functions appropriate to our
particular problem.
Instead of the block diagram algebra, we could have combined the Laplace
domain equations of Section 4.15 directly, eliminating intermediate
variables until we arrived at (4.16-4).
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Furthermore, we could have proceeded entirely in the time domain, as we
did in Lesson 3. That is, the second-order process ODE could have been
combined with the first-order valve ODE and algebraic equations for
sensor and controller to arrive at an ODE for the controlled variable with
disturbance and set point forcing functions.
We have used new tools - the Laplace transform and the block diagram but the underlying objective, and the relationships between inputs and
outputs, were the same as working in the time domain. This is not
mysterious.
4.18 calculating closed-loop responses
But how does the closed-loop perform? We approach this by simplifying
(4.16-4).
τ vs + 1
τ1τ 2 τ v
C 'Ai (s)
C 'A 2 (s) =
⎛ τ + τ + τ ⎞ 1 + K v K c K s C Ac F
⎛1 1 1⎞
s 3 + ⎜⎜ + + ⎟⎟s 2 + ⎜⎜ 1 2 v ⎟⎟s +
τ1τ 2 τ v
⎝ τ1τ 2 τ v ⎠
⎝ τ1 τ 2 τ v ⎠
1 + K v K c K s C Ac F
τ1τ 2 τ v
C 'A 2,sp (s)
+
⎛ τ + τ + τ ⎞ 1 + K v K c K s C Ac F
⎛1 1 1⎞
s 3 + ⎜⎜ + + ⎟⎟s 2 + ⎜⎜ 1 2 v ⎟⎟s +
τ1τ 2 τ v
⎝ τ1τ 2 τ v ⎠
⎝ τ1 τ 2 τ v ⎠
(4.18-1)
Clearly we did not succeed at that... As with (4.5-4) and (4.11-3), we
would like to substitute a particular disturbance for C′Ai(s) in (4.18-1) and
invert the result to obtain the time response of CA2. Here we encounter an
obstacle: our transform tables do not feature anything as complicated as
(4.18-1). Furthermore, to use partial fraction expansion we must find the
roots of the cubic equation; however, we are unlikely to find an analytical
expression for these. That is unfortunate, because it would be helpful to
know how the transfer function poles depend on the controller gain Kc.
We resort to numerical methods. Here is our plan:
• do a partial-fraction expansion of each transfer function in (4.18-1) in
terms of the poles
• multiply each term in the expansion by the disturbance of interest and
invert to find the response
• for a particular value of controller gain Kc, find the poles numerically
• repeat for different values of Kc to map out the behavior
This expedient of using numerical calculations does not show us the
functional dependence of the response on the parameters, but it does get
the job done.
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We will begin with the disturbance transfer function in (4.18-1). The
polynomial ratio part of the transfer function is written as the sum of
fractions.
τ vs + 1
⎛ τ + τ + τ ⎞ 1 + K v K c K s C Ac F
⎛1 1 1⎞
s 3 + ⎜⎜ + + ⎟⎟s 2 + ⎜⎜ 1 2 v ⎟⎟s +
τ1τ 2 τ v
⎝ τ1τ 2 τ v ⎠
⎝ τ1 τ 2 τ v ⎠
τ vs + 1
=
(s − α1 )(s − α 2 )(s − α 3 )
=
(4.18-2)
C1
C2
C3
+
+
s − α1 s − α 2 s − α 3
The poles of the transfer function are αi, and the coefficients Ci depend on
these poles, as well as the numerator. We will keep in mind that both αi,
and Ci depend on the system time constants and gains, as well as the
controller gain Kc. Solving for the coefficients is an algebra problem. The
results are
C1 =
τ v α1 + 1
(α1 − α 2 )(α1 − α 3 )
C2 =
τvα 2 + 1
(α 2 − α1 )(α 2 − α 3 )
C3 =
τvα3 + 1
(α 3 − α1 )(α 3 − α 2 )
(4.18-3)
Thus numerical values of coefficients Ci can be computed for each set of
poles αi.
Now we use expansion (4.18-2) to rewrite (4.18-1) for a disturbance
response. With no change in set point, C′A2,sp(s) is identically zero.
⎡ C1
C3 ⎤ '
C2
+
+
⎥ C Ai (s)
⎢
−
α
−
α
−
α
s
s
s
1
2
3⎦
⎣
−1
−1
⎡ −1
⎤
C3 ⎥
C1
C2
⎢
α
1
α
α
⎢ 1
⎥ C 'Ai (s)
=
+ 2
+ 3
τ1τ 2 τ v ⎢ − 1 s + 1 − 1 s + 1 − 1 s + 1⎥
⎢⎣ α1
⎥⎦
α3
α2
C 'A 2 (s) =
1
τ1τ 2 τ v
(4.18-4)
Now, as an example, we pose a step disturbance in the inlet composition.
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C 'Ai ( t ) = ΔCU ( t − t d )
C 'Ai (s) =
(4.18-5)
ΔC − t ds
e
s
We substitute the step disturbance (4.18-5) into the system model (4.18-4)
to obtain
⎡ −1
⎤
−1
−1
C3 ⎥
C1
C2
⎢
α3
ΔC ⎢ α1
α2
⎥ e − t ds
C 'A 2 (s) =
+
+
⎢
τ1τ 2 τ v ⎛ − 1
⎞ ⎥
⎞ ⎛ −1
⎞ ⎛ −1
⎢ ⎜⎜ s + 1⎟⎟s ⎜⎜ s + 1⎟⎟s ⎜⎜ s + 1⎟⎟s ⎥
⎠ ⎝ α3
⎠ ⎝ α2
⎠ ⎦⎥
⎣⎢ ⎝ α1
(4.18-6)
Each term inverts to a step response. Remembering to apply the time
delay, we find the result
C′A 2 (t) =
⎤
ΔCU(t − t d ) ⎡ −C1
− C3
−C 2
1 − eα1 (t − t d ) +
1 − eα2 (t − t d ) +
1 − eα3 (t − t d ) ⎥
⎢
τ1τ2 τ v
α2
α3
⎣ α1
⎦
(
)
(
)
(
)
(4.18-7)
Examining (4.18-7) we learn that the exponential terms contribute
according to the magnitude of the pole αi: small poles (larger time
constants) cause the term to persist. We see that there will be offset,
because C′A2 does not go to zero at long times. The amount of offset will
depend on the magnitude of the coefficients Ci; our experience in Lesson 3
would suggest that these will become smaller as the controller gain Kc
increases.
4.19 calculating the response for a particular example
We begin with similar parameter values to our example in Lesson 3:
F = 1.2 m3 min-1
Fc,r = 6×10-3 m3 min-1
V1 = 6 m3 (thus τ1 = 5 min)
V2 = 4 m3 (thus τ2 = 3.33 min)
CAi,r = 8 kg m-3
CAo,r = 10 kg m-3
CAc = 400 kg m-3
τv = 0.1 min
Kv = 0.01 m3 min-1 controller_out-1
Ks = 0.5 controller_in m3 kg-3
We may calculate roots in (4.18-2) with calculators, spreadsheets, or
computer code. For example, using matlab we obtain roots of polynomial
s2 + 3s +4 by
>> roots ([1 3 4])
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ans =
-1.5000 + 1.3229i
-1.5000 - 1.3229i
Table 4.19-1: poles of closed loop disturbance transfer function
Kc (out in-1)
α1 (min-1) α2 (min-1)
α3 (min-1)
0
-10
-0.2
-0.3
0.012
-10.0001
-0.2143
-0.2856
0.024
-10.0003
-0.2437
-0.2561
Notice that zero controller gain leads to poles equal to the negative inverse
of the three system time constants. Thus our closed-loop transfer function
reduces to describe the behavior of the process alone, under open-loop
conditions. After using the poles in Table 4.19-1 to compute the solution
(4.18-7) we obtain a plot of the response behavior. Indeed controller gain
can be increased to reduce the effects of the input disturbance.
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1.5
1
0.5
0
-0.5
00
10
10
20
20
0.0
30
30
40
40
30
40
Kc = 0
-0.5
-6
3 -1
make-up deviation (10 m s )
inlet comp deviation (kg m
-3
)
Lesson 4: Two Tanks in Series
-1.0
0.012
-1.5
-2.0
0.024
-2.5
-3
outlet composition deviation (kg m)
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0
10
20
time (min)
Figure 4.19-1: Step response of two tanks under proportional control
4.20 surprise - increasing gain introduces oscillations!
We have been very tentative with the gain setting, so we act more
aggressively to suppress offset. The poles become complex!
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Table 4.20-1: complex poles of transfer function
Kc (out in-1)
α1 (min-1)
α2 (min-1)
0.1
-10.001
-0.249 + j0.088
10
-10.10
-0.198 + j1.00
50
-10.48
-0.011 + j2.20
α3 (min-1)
-0.249 - j 0.088
-0.198 - j1.00
-0.011 - j2.20
We must modify (4.18-7) to accommodate complex numbers. The roots
α2 and α3 are complex conjugates, and (recalling our discussion of partial
fraction expansion) so are coefficients C2 and C3. We define four new real
quantities to replace the complex ones:
α 2 = a + jb
α 3 = a − jb
C2
= A + jB
α2
C3
= A − jB
α3
(4.20-1)
We substitute these definitions into (4.18-7), recalling Euler’s relation,
e (a + jb )t = e at (cos bt + j sin bt )
(4.20-2)
to obtain
C 'A 2 ( t ) =
⎤
ΔCU ( t − t d ) ⎡ − C1
1 − e α1 ( t − t d ) − 2A + 2e a ( t − t d ) (A cos b( t − t d ) − B sin b( t − t d ) )⎥ (4.20-3)
⎢
τ1τ 2 τ v
⎣ α1
⎦
(
)
Parameters A, B, a, and b are not variables with time, in the sense of CA2,
but they do depend on the value of the controller gain Kc. Parameters a
and b are found (via the root-finding procedure) in Table 4.20-1. A and B
come via complex algebra from (4.18-3).
A=
(
)
− bτ v a 2 + b 2 + α1b − 2ab
2
2b a 2 + b 2 (α1 − a ) + b 2
(
τ (a
B=
v
)[
)(α
]
2
+b
1 − a ) + a (α1 − a ) + b
2
2b a 2 + b 2 (α1 − a ) + b 2
2
2
(
)[
(4.20-4)
]
Taking data from Table 4.20-1 and using (4.20-4), we can plot response
(4.20-3).
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Process Dynamics, Operations, and Control
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-5
3 -1
make-up flow deviation (10 m s )
-3
inlet comp deviation (kg m)
Lesson 4: Two Tanks in Series
1.5
1
0.5
0
-0.5
00
10
10
20
20
30
30
4400
0
-2
-4
-6
-8
-10
-12
-14
-3
outlet composition deviation (kg m)
0.90
0.80
Kc = 0.1
0.70
0.60
0.50
0.40
0.30
0.20
0.10
10
50
0.00
0
10
20
30
40
time (min)
Figure 4.20-1: Oscillatory step response
For the single tank of Lesson 3, the closed loop behavior was qualitatively
the same as that of the process itself. Here, however, closing the loop has
introduced behavior we would NOT see in the process alone: the response
variable oscillates in response to a steady input. The key is the third-order
characteristic equation, which can admit complex roots. The transfer
function is third order because the second-order process was placed in a
feedback loop with a first-order valve. If the system mathematics provide
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Lesson 4: Two Tanks in Series
a true representation of the process equipment, we will see oscillations in
operation.
4.21 closed-loop stability by the Bode criterion
Figure 4.20-1 shows us that, despite the onset of oscillations, increasing
gain suppresses the offset. However, we must still be mindful of the
possibility of instability. In Figure 4.21-1 we plot the poles on a complex
plane. We see that increasing gain brings the closed loop to the point of
oscillation and then increases the imaginary (oscillatory) component. We
observe also that the real component approaches zero. Recalling that
stability depends on the real parts of the poles being negative, we see that
the closed loop will become unstable at a controller gain between 50 and
60.
4.00
3.00
arrows show increasing Kc
-1
imaginary (min )
2.00
from 0 to 100
1.00
0.00
-1.00
-2.00
-3.00
-4.00
-12
-10
-8
-6
-4
-2
0
2
-1
real (min )
Figure 4.20-1: Root locus plot
This root locus plot provides a map of the stability limit; this is
particularly helpful, because we were unable to derive a single expression
that showed the effect of controller gain on the real parts of the poles. We
might imagine that finding the poles will only become more difficult as we
consider more complicated processes and controllers. Hence we introduce
an alternative means of predicting stability: the Bode criterion.
We develop the criterion intuitively; recalling Figure 4.14-1, we begin by
realizing that instability in a previously stable system happens because of
feedback in a closed loop. Suppose that the output signal y′c contains
some fluctuating component at a particular frequency ωc. This component
is inverted in the comparator by being subtracted from the set point, and is
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Lesson 4: Two Tanks in Series
fed to the controller. If the loop (process, controller, sensor, valve, etc.)
contributes a phase delay of -180° at frequency ωc, the inverted signal
returns to the output in phase to reinforce the fluctuation. If in addition
the amplitude ratio at ωc is greater than one, the fluctuation will grow.
We are not contriving a circumstance here. In any realistic process there
will be small disturbances, fluctuations over a wide domain of frequency.
The loop processes these fluctuations according to its frequency response
characteristics. Depending on the amplitude ratio, signals at some
frequencies may be amplified. Assuredly, signals at high frequencies will
be delayed by at least 180°. If these conditions overlap, there will be an
oscillating signal that will grow. We need not supply it; the system will
select it from the spectrum of background noise. This intuitive
development is described in more detail by Marlin (Sec.10.6).
To turn our intuition into a method, we return to the general description of
the closed loop transfer function in (4.16-2). Recall that the poles are the
roots of the characteristic equation, which is the denominator of the
transfer function. This characteristic equation is always of the form 1 plus
the product of the transfer functions around the loop. For convenience, we
will call this product the loop transfer function GL.
characteristic equation = 1 + G s G c G v G m
= 1+ GL
(4.21-1)
It is the amplitude ratio and phase angle, that is, the frequency response, of
GL that determines whether signals will grow in the loop. First we find ωc,
the frequency at which the phase delay is -180°. At this crossover
frequency, we inspect the amplitude ratio; if it is less than one, the system
will attenuate reinforced disturbances, and thus be stable.
Thus the Bode criterion evaluates the stability of (1 + GL)-1 from the
frequency response of GL.
For our process, (4.16-4) gives
G L (s) =
C Ac F
Kv
KK
(τ1s + 1)(τ2s + 1) τ vs + 1 c s
(4.21-2)
The phase angle of GL is the sum of the phase angles of the various
elements in (4.21-2).
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Lesson 4: Two Tanks in Series
∠G L ( jω) = ∠
C Ac
1
1
1
KcKsK v + ∠
+∠
+∠
F
τ1s + 1
τ 2s + 1
τ vs + 1
= 0 + tan
−1
(− ωτ1 ) + tan (− ωτ2 ) + tan (− ωτ v )
−1
(4.21-3)
−1
Equation (4.21-3) may be solved for the crossover frequency ωc; that is,
the frequency at which the loop delays the signal by -180º.
− 180D = tan −1 (− ωc τ1 ) + tan −1 (− ωc τ 2 ) + tan −1 (− ωc τ v )
(4.21-4)
The amplitude ratio of GL is the product of the amplitude ratios of the
various elements in (4.21-2).
G L ( jω) =
C Ac
1
1
1
KcKsK v ×
×
×
F
τ1s + 1 τ 2s + 1 τ vs + 1
C
= Ac K c K s K v
F
1
1
1 + ω2 τ1
2
(4.21-5)
1
1 + ω2 τ 2
2
1 + ω2 τ v
2
We are particularly interested in the amplitude ratio at the crossover
frequency, RAc.
R Ac =
C Ac
KcKsK v
F
1
1
2
1 + ωc τ1
2
1
2
1 + ωc τ 2
2
2
1 + ωc τ v
2
(4.21-6)
Using the data in Section 4.19, we find the crossover frequency from
(4.21-4) to be 2.25 radians minute-1. We notice that phase lag in the loop
depends only on the tanks and valve; the proportional controller, being a
“pure gain” system, contributes no lag to the dynamic response of the
loop. Hence the crossover frequency does not vary with the controller
gain setting.
Using the crossover frequency and further data from Section 4.19, we find
from (4.21-6) that the crossover amplitude ratio will be 1 when the
controller gain Kc is 52.55. The effect of controller gain is to amplify the
signals in the loop. Around Kc = 52.55, therefore, the system output will
oscillate unabated at frequency ωc. At higher gain settings, the amplitude
of the oscillation will grow in time. (The frequency of these oscillations
depends on the poles of the transfer function.)
Figure 4.21-1 is a Bode plot for the loop transfer function GL, showing
gains below, at, and above the instability threshold. The stability
threshold (amplitude ratio = 1, phase angle = -180°) is shown by a single
point at the crossover frequency
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Lesson 4: Two Tanks in Series
1000
amplitude ratio
100
10
1
Kc = 100
52.55
0.1
0.01
0.01
0
25
0.1
1
10
100
phase angle (deg)
-50
-100
-150
crossover point
-200
-250
-300
0.01
0.1
1
10
100
-1
ω (radians min )
Figure 4.21-1: Bode plot for loop transfer function
Figure 4.21-2 shows unstable step responses at gains of 60 and 100. The
latter response quickly gets out of hand. Notice how the make-up flow
varies in response to the increasing error in the outlet composition.
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-3
-5 3 -1
make-up flow deviation (10 m s ) inlet comp deviation (kg m)
Lesson 4: Two Tanks in Series
1.5
1
0.5
0
-0.5
00
55
10
10
15
15
20
20
25
25
30
30
50
40
30
20
10
0
-10
-20
-30
-40
-50
-3
outlet composition deviation (kg m)
0.30
0.20
Kc = 100
0.10
60
0.00
-0.10
-0.20
-0.30
0
5
10
15
20
25
30
time (min)
Figure 4.21-2: Step response for two-tank process at high gains
4.22 tuning based on stability limit - gain and phase margin
We tune a controller seeking good performance, somewhere between the
extremes of “no control” and “instability”. One method of tuning is
simply to maintain a reasonable distance from the instability limit and
presume that the result is an improvement over having no control. Thus,
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Lesson 4: Two Tanks in Series
we find the instability threshold and tune the controller to leave margins
between these conditions and normal operation. The margins are
indicated in Figure 4.22-1, which shows a Bode plot for the loop transfer
function GL at some arbitrary controller setting.
RA
(2)
1
GM > 1
RAc
(3)
φ
φ1
PM > 0
(1)
-180
(4)
ω1
ωc
ω
Figure 4.22-1: Illustration of gain margin and phase margin at a
single controller setting
The gain margin and phase margin are the distances shown on the
ordinates. Their definitions are
GM =
1
R Ac
(> 1 for stability )
PM = 180D + φ1
(> 0 for stability )
(4.22-1)
The controller setting determines the amplitude ratio and phase angle
curves. From those curves we then calculate the margins to see if they are
satisfactory:
(1)
(2)
(3)
(4)
use a phase angle of -180° to find the crossover frequency ωc
use an amplitude ratio of 1 to find the frequency ω1
use ωc to find the amplitude ratio RAc, and thus GM
use ω1 to find the phase angle φ1, and thus PM
In Figure 4.22-1, the system is stable. However, as the controller gain is
increased, the Bode plot will shift so that ω1 and ωc approach each other.
At the instability threshold, ω1 equals ωc, the gain margin is 1, and phase
margin is zero.
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Lesson 4: Two Tanks in Series
A procedure for tuning proportional controllers by stability margin is:
(1)
(2)
(3)
(4)
(5)
use a phase angle of -180° to find the crossover frequency ωc
at ωc, find the gain that makes RAc = 1 (stability limit)
at ωc, reduce the gain to make RAc = 1/GM (gain margin)
use a phase angle of PM -180° to find the frequency ω1
at ω1, find the gain that makes RA = 1 (phase margin)
Marlin (Sec.10.8) recommends tuning to maintain GM ~ 2 and PM ~ 30°.
Typically one or the other will be limiting.
Figure 4.22-2 shows the results of calculations for our tank example.
Earlier, we found the crossover frequency from (4.21-4). Then (4.21-6)
was solved for the controller gain that gave RAc = 0.5 (thus GM = 2). The
result was Kc = 26.3. Then (4.21-3) was solved for the frequency ω1 to
give a phase angle of -150° (thus PM = 30°). Then (4.21-5) was solved
for the controller gain that gave an amplitude ratio of 1 at frequency ω1.
The result was a much lower gain of 7. Therefore for our system, PM is
limiting, and the lower gain would be chosen. For reference, Figure 4.222 also shows the stability limit determined earlier.
The gain and phase margins have given us a tuning criterion for selecting
a controller gain. Using the chosen gain, we can now predict the
performance in response to disturbances and set point changes. The
calculations would be similar to those illustrated in Sections 4.19 and
4.20: a partial fraction expansion leading to an expression for the
response, with parameter values based on numerical root-finding.
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Lesson 4: Two Tanks in Series
amplitude ratio
100
10
Kc = 52.55
1
26.3
0.1
7
0.01
0.01
0
0.1
1
10
100
0.1
1
10
100
phase angle (deg)
-50
-100
-150
-200
-250
-300
0.01
-1
ω (radians min )
Figure 4.22-2: Bode plot illustrating GM and PM limits on gain
4.23 conclusion
We have completed dynamic analysis and control of a more complicated
process than in Lesson 3. In doing so we have introduced new tools for
analysis - the Laplace transforms and block diagrams - and developed
stability and tuning criteria.
Was it a good idea to listen to the appealing advisor and put the make-up
flow into the first tank? A good way to examine the question would be to
repeat the full analysis for the other case. Even without doing that,
however, we might reflect how removing one lag from the system might
affect the Bode stability criterion for the closed loop…
4.24 reference
Marlin, Thomas E. Process Control. 2nd ed. Boston, MA: McGraw-Hill, 2000.
ISBN: 0070393621.
4.25
a
nomenclature
constant
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Lesson 4: Two Tanks in Series
A
A1,A2
b
B
C
C1…
CA1
CA2
CAc
CAi
CAs
ΔC
F
Fc
f
G
Im
j
K
L
N(s)
r
Re
RA
RAc
s
t
td
U
V1
V2
x(t)
y(t)
α1 …
ε
τ1
τ2
τv
ξ
ω
ωc
constant
constants of integration
constant
constant
constant
constants in partial fraction expansion
intermediate stream concentration of solute A
exit stream concentration of solute A
make-up stream concentration of solute A
inlet stream concentration of solute A
reference concentration of solute A at steady state
change in solute concentration
volumetric flowrate
volumetric flowrate of make-up stream
function
transfer function
operator that takes imaginary part of complex number
square root of -1
gain (time-independent part of transfer function)
Laplace operator
polynomial in s
dummy variable in polynomial characteristic equation
operator that takes real part of complex number
the amplitude ratio of the loop transfer function
the amplitude ratio of the loop transfer function at the crossover frequency
complex Laplace domain variable
time
time at which disturbance occurs
unit step function
volume of tank 1
volume of tank 2
input signal to system
output signal from system
roots of polynomial in s
error; set point minus controlled variable
time constant of tank 1
time constant of tank 2
time constant of valve
dummy variable of integration
radian frequency (has dimensions of radians time-1)
crossover frequency, at which loop transfer function lag is -180°
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